JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 16)
In a Young’s double slit experiment, 16 fringes
are observed in a certain segment of the
screen when light of a wavelength 700 nm is
used. If the wavelength of light is changed to
400 nm, the number of fringes observed in the
same segment of the screen would be
28
24
30
18
Explanation
Let the length of segment is "$$l$$"
Let N is the no. of fringes in "$$l$$" and w is fringe width.
$$ \therefore $$ Nw = $$l$$
$$ \Rightarrow $$ N$$\left( {{{\lambda D} \over d}} \right)$$ = $$l$$
As in both cases segment length is same.
$$ \therefore $$ $${{{{N_1}{\lambda _1}D} \over d} = l}$$
and $${{{{N_2}{\lambda _2}D} \over d} = l}$$
$$ \therefore $$ $${{{{N_1}{\lambda _1}D} \over d}}$$ = $${{{{N_2}{\lambda _2}D} \over d}}$$
$$ \Rightarrow $$ $${{N_1}{\lambda _1}}$$ = $${{N_2}{\lambda _2}}$$
$$ \Rightarrow $$ 16 × 700 = N2 × 400
$$ \Rightarrow $$ N2 = 28
Let N is the no. of fringes in "$$l$$" and w is fringe width.
$$ \therefore $$ Nw = $$l$$
$$ \Rightarrow $$ N$$\left( {{{\lambda D} \over d}} \right)$$ = $$l$$
As in both cases segment length is same.
$$ \therefore $$ $${{{{N_1}{\lambda _1}D} \over d} = l}$$
and $${{{{N_2}{\lambda _2}D} \over d} = l}$$
$$ \therefore $$ $${{{{N_1}{\lambda _1}D} \over d}}$$ = $${{{{N_2}{\lambda _2}D} \over d}}$$
$$ \Rightarrow $$ $${{N_1}{\lambda _1}}$$ = $${{N_2}{\lambda _2}}$$
$$ \Rightarrow $$ 16 × 700 = N2 × 400
$$ \Rightarrow $$ N2 = 28
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